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Can anyone help me prove this? This one is from Malik's Fundamentals of Abstract Algebra.

An ideal $I$ of a ring $R$ is called a minimal ideal if $I≠{0}$ and there does not exist any ideal $J$ of R such that ${0}≠J⊂I$.

If $I$ is a minimal ideal of a commutative ring $R$ with $1$, show that either $I^2={0}$ or $I=eR$ for some idempotent $e∈R$.

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up vote 6 down vote accepted

Suppose $I^2 \neq 0$, then $\exists a\in I$ such that $aI \neq 0$. Hence, $aI = I$, and so $\exists e\in I$ such that $ae = a$.

Now, $J = (e^2-e)R \subset I$ and $J\neq I$ (because if $J=I$, then $aI = a(e^2-e)R = 0$), and hence $J = 0$, whence $e$ is an idempotent and $I = eR$.

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There is a detail I am missing. Could you please explain why $a(e^2−e)R=0$. – Rodney Coleman Dec 29 '13 at 16:25
@RodneyColeman $a(e^2-e)=aee-ae=ae-ae=0$. – user114539 Dec 29 '13 at 16:34
Many thanks. I should have seen it. – Rodney Coleman Dec 31 '13 at 9:35
@Prahlad could you explain how $J=(e^2-e)R$ is a subset of $I$ – zed111 Apr 27 '15 at 17:47

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